Description

In middle school, teachers used to encourage us to pick up pretty sentences so that we could apply those sentences in our own articles. One of my classmates ZengXiao Xian, wanted to get sentences which are different from that of others, because he thought the distinct pretty sentences might benefit him a lot to get a high score in his article.
Assume that all of the sentences came from some articles. ZengXiao Xian intended to pick from Article A. The number of his classmates is n. The i-th classmate picked from Article Bi. Now ZengXiao Xian wants to know how many different sentences she could pick from Article A which don't belong to either of her classmates?Article. To simplify the problem, ZengXiao Xian wants to know how many different strings, which is the substring of string A, but is not substring of either of string Bi. Of course, you will help him, won't you?

solution

以A建立SAM
在字符串中一个节点 \(p\) 可以接受的子串为 \(len[p]\)，但是有一些被父亲节点接受了，所以他实际只可以接受 \(len[p]-len[fa[p]]\)，所以我们拿 \(bi...bn\) 在上面跑，记录一个在一个节点最多被匹配的长度 \(f[i]\)，每一个节点的贡献就是 \(Max(len[i]-Max(f[i],len[fa[i]]),0)\)，注意 \(f[i]\) 要向父亲上传

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define RG register
#define il inline
#define iter iterator
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
const int N=200005;
char s[N];int m,n,cnt=1,len[N],ch[N][27],p,cur=1,fa[N];
void build(int c,int id){
   p=cur;cur=++cnt;len[cur]=id;
   for(;p && !ch[p][c];p=fa[p])ch[p][c]=cur;
   if(!ch[p][c])fa[cur]=1;
   else{
      int q=ch[p][c];
      if(len[p]+1==len[q])fa[cur]=q;
      else{
         int nt=++cnt;len[nt]=len[p]+1;
         memcpy(ch[nt],ch[q],sizeof(ch[q]));
         fa[nt]=fa[q];fa[q]=fa[cur]=nt;
         for(;ch[p][c]==q;p=fa[p])ch[p][c]=nt;
      }
   }
}
int kase=0,f[N];
void matcher(){
   int c,l=strlen(s+1),p=1,x=0;
   for(int i=1;i<=l;i++){
      c=s[i]-'a';
      if(!ch[p][c]){
         while(p && !ch[p][c])p=fa[p];
         if(!p)p=1,x=0;
         else x=len[p]+1,p=ch[p][c];
      }
      else x++,p=ch[p][c];
      if(x>f[p])f[p]=x;
   }
}
int c[N],sa[N];
void Clear(){
   memset(c,0,sizeof(c));cnt=1;cur=1;
   memset(ch,0,sizeof(ch));memset(f,0,sizeof(f));
   memset(fa,0,sizeof(fa));
}
void work()
{
   Clear();
   scanf("%d%s",&m,s+1);
   n=strlen(s+1);
   for(int i=1;i<=n;i++)build(s[i]-'a',i);
   for(int i=1;i<=m;i++){
      scanf("%s",s+1);
      matcher();
   }
   for(int i=1;i<=cnt;i++)c[len[i]]++;
   for(int i=1;i<=n;i++)c[i]+=c[i-1];
   for(int i=cnt;i;i--)sa[c[len[i]]--]=i;
   for(int i=cnt,x;i;i--){
      x=sa[i];
      f[fa[x]]=Max(f[fa[x]],f[x]);
   }
   long long ans=0,tmp;
   for(int i=1;i<=cnt;i++){
      tmp=Max(f[i],len[fa[i]]);
      ans+=Max(len[i]-tmp,0);
   }
   printf("Case %d: %lld\n",++kase,ans);
}

int main()
{
	int T;cin>>T;
        while(T--)work();
	return 0;
}